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Rational function

Module by: Sunil Kumar Singh

Rational function is defined in similar fashion as rational number is defined in terms of numerator and denominator. Implicitly, we refer “real” rational function here. It is defined as the ratio of two real polynomials with the condition that polynomial in the denominator is not a zero polynomial.

f x = p x q x ; q x ≠ 0 f x = p x q x ; q x ≠ 0

Rational function is not defined for values of x for which denominator polynomial evaluates to zero as ratio “p(x)/0” is not defined. Some examples of rational function are :

f x = 2 x 2 − x + 1 2 x 2 − 5 x − 3 ; x ≠ - 1 2 , x ≠ 3 f x = 2 x 2 − x + 1 2 x 2 − 5 x − 3 ; x ≠ - 1 2 , x ≠ 3

g x = x + 1 2 x 2 − x + 1 g x = x + 1 2 x 2 − x + 1

h x = 2 x 4 − x 2 + 1 x + 1 ; x ≠ - 1 h x = 2 x 4 − x 2 + 1 x + 1 ; x ≠ - 1

Note second example function, g(x) above. There is no exclusion point for this rational polynomial. The denominator polynomial is 2 x 2 − x + 1 2 x 2 − x + 1 , whose determinant is negative and coefficient of x 2 x 2 term is positive. It means denominator of g(x) is positive for all values of x. We should also note that values of x being excluded are points - not a continuous interval. Further, the notation to denote exclusion is an “inequation” – not “inequality” – because notation x ≠ - 1 x ≠ - 1 negates corresponding equation x = -1. Recall that inequality, on the other hand, compares relative values.

Domain of rational function

Domain of rational function is domain of numerator polynomial minus exclusion points as determined by zeroes of denominator polynomial. Since domain of polynomial is R, domain of rational polynomial is R minus exclusion points determined by denominator. The domains for three rational functions given above are :

Domain of f(x) = R − { − 1 2 , 3 } Domain of f(x) = R − { − 1 2 , 3 } Domain of g(x) = R Domain of g(x) = R Domain of h(x) = R − { − 1 } Domain of h(x) = R − { − 1 }

Real values of rational function

In order to find real values of rational function for real x, we rearrange the given rational function to form a quadratic equation in x. Let us consider a rational polynomial of degree 2 given by (we mean that the highest degree of polynomials involved in the ratio is 2) :

y = f x = a 1 x 2 + b 1 x + c 1 a 2 x 2 + b 2 x + c 2 y = f x = a 1 x 2 + b 1 x + c 1 a 2 x 2 + b 2 x + c 2

Rearranging to form a quadratic equation in x, we have :

⇒ y a 2 x 2 + b 2 x + c 2 = a 1 x 2 + b 1 x + c 1 ⇒ y a 2 x 2 + b 2 x + c 2 = a 1 x 2 + b 1 x + c 1

⇒ y a 2 − a 1 x 2 + y b 2 − b 1 x + c 2 y − c 1 = 0 ⇒ y a 2 − a 1 x 2 + y b 2 − b 1 x + c 2 y − c 1 = 0

For x to be real, D≥0. Hence,

⇒ D = y b 2 − b 1 2 − 4 x y a 2 − a 1 c 2 y − c 1 ≥ 0 ⇒ D = y b 2 − b 1 2 − 4 x y a 2 − a 1 c 2 y − c 1 ≥ 0

We see that discriminant itself is a quadratic inequality. Depending on the nature of coefficient of “ y 2 y 2 ” in the quadratic equation and determinant of the corresponding quadratic equation, the inequality is solved for “y”. This, in turn, allows us to determine the real interval(s) of y corresponding to real x.

We should clearly understand that these are real values of y corresponding to real x. This interval need not be the range of the function. Recall that range of a function contains values of y for values of x in the domain of function – not all real values of x. Now function is not defined for certain values of x, which are zeroes of denominator of the rational function. Hence domain is not the real number set R. This distinction should always be kept in mind.

Example 1

Problem : Find value of x for which given function has least value.

y = f x = x 2 − 6 x + 5 x 2 + 2 x + 1 y = f x = x 2 − 6 x + 5 x 2 + 2 x + 1

Solution : Rearranging to form a quadratic equation in x, we have :

⇒ y x 2 + 2 y x + y = x 2 − 6 x + 5 ⇒ y x 2 + 2 y x + y = x 2 − 6 x + 5 ⇒ y − 1 x 2 + 2 y + 3 x + y − 5 = 0 ⇒ y − 1 x 2 + 2 y + 3 x + y − 5 = 0

For x real, D≥0.

⇒ 4 y + 3 2 − 4 y − 1 X y − 5 ≥ 0 ⇒ 4 y + 3 2 − 4 y − 1 X y − 5 ≥ 0 ⇒ y 2 + 6 y + 9 − y 2 − 6 y + 5 ≥ 0 ⇒ y 2 + 6 y + 9 − y 2 − 6 y + 5 ≥ 0 ⇒ 12 y + 4 ≥ 0 ⇒ 12 y + 4 ≥ 0 ⇒ y ≥ − 1 3 ⇒ y ≥ − 1 3

The real values of y, therefore, lies in the interval [-1/3, ∞). The least value of y in the interval of real values is -1/3. We should, however, check that value of y=-1/3 does not correspond to value of x which is not permitted. Here, the denominator polynomial is x 2 + 2 x + 1 = x + 1 2 x 2 + 2 x + 1 = x + 1 2 . Thus, x ≠ - 1 x ≠ - 1 . The domain of the function is R-{-1}.

Now, we calculate value of x corresponding to y as :

⇒ − 1 3 = x 2 − 6 x + 5 x 2 + 2 x + 1 ⇒ − 1 3 = x 2 − 6 x + 5 x 2 + 2 x + 1

⇒ x 2 − 4 x + 4 = 0 ⇒ x 2 − 4 x + 4 = 0 ⇒ x − 2 2 = 0 ⇒ x − 2 2 = 0 ⇒ x = 2 ⇒ x = 2

This point belongs to the domain of the function as it is different to excluded value. Hence, least value of y is -1/3.

Example 2

Problem : For what values of “a”, the function given here assumes all real values for real x.

y = f x = a x 2 + 3 x − 4 3 x − 4 x 2 + a y = f x = a x 2 + 3 x − 4 3 x − 4 x 2 + a

Solution : Rearranging to form a quadratic equation in x, we have :

⇒ 3 y x − 4 y x 2 + a y = a x 2 + 3 x − 4 ⇒ 3 y x − 4 y x 2 + a y = a x 2 + 3 x − 4

⇒ a + 4 y x 2 + 3 1 − y x − a y + 4 = 0 ⇒ a + 4 y x 2 + 3 1 − y x − a y + 4 = 0

For x real, D≥0.

⇒ 9 1 − y 2 + 4 a + 4 y X a y + 4 ≥ 0 ⇒ 9 1 − y 2 + 4 a + 4 y X a y + 4 ≥ 0

⇒ 9 + 16 a y 2 + 4 a 2 + 46 y + 9 + 16 a ≥ 0 ⇒ 9 + 16 a y 2 + 4 a 2 + 46 y + 9 + 16 a ≥ 0

This is a quadratic inequality in y. This inequality holds when coefficient of “ y 2 y 2 ” is positive and discriminant is less than equal to zero i.e. D’≤0.

⇒ 9 + 16 a > 0 ⇒ 9 + 16 a > 0

⇒ a > - 9 16 ⇒ a > - 9 16

and

D ′ = 4 a 2 + 46 2 − 4 9 + 16 a 2 ≤ 0 D ′ = 4 a 2 + 46 2 − 4 9 + 16 a 2 ≤ 0 ⇒ 2 a 2 + 23 2 − 9 + 16 a 2 ≤ 0 ⇒ 2 a 2 + 23 2 − 9 + 16 a 2 ≤ 0 ⇒ 2 a 2 + 23 + 9 + 16 a 2 a 2 + 23 − 9 − 16 a ≤ 0 ⇒ 2 a 2 + 23 + 9 + 16 a 2 a 2 + 23 − 9 − 16 a ≤ 0 ⇒ 2 a 2 + 16 a + 32 2 a 2 − 16 a + 14 ≤ 0 ⇒ 2 a 2 + 16 a + 32 2 a 2 − 16 a + 14 ≤ 0 ⇒ a 2 + 8 a + 16 a 2 − 8 a + 7 ≤ 0 ⇒ a 2 + 8 a + 16 a 2 − 8 a + 7 ≤ 0 ⇒ a + 4 2 a 2 − 8 a + 7 ≤ 0 ⇒ a + 4 2 a 2 − 8 a + 7 ≤ 0 ⇒ a 2 − 8 a + 7 ≤ 0 ⇒ a 2 − 8 a + 7 ≤ 0 ⇒ a 2 − 8 a + 7 ≤ 0 ⇒ a 2 − 8 a + 7 ≤ 0

The roots of corresponding quadratic equation in a is :

⇒ a = 1,7 ⇒ a = 1,7

Hence interval of a satisfying inequality is [1,7]

We have two intervals i.e. a>-9/16 and ⇒ a ∈ [ 1,7 ] ⇒ a ∈ [ 1,7 ] corresponding to two simultaneous conditions. Therefore, values of a is intersection of these two intervals :

⇒ a ∈ [ 1,7 ] ⇒ a ∈ [ 1,7 ]

Range of rational function

The set of real values of rational polynomial for real values of x need not be the range of the function. It is because rational function is not defined for zeroes of polynomial in denominator. In previous section, we evaluated values of function for real x. But, domain may not be the real number set, but subset of R, which excludes certain values of x. We need to exclude values of “y”, which corresponds to values of x for which denominator becomes zero. This statement, however, is slightly confusing, because function is not defined for those values of x in the first place. How would we determine values of y corresponding to values of x for which function reduces to indeterminate form involving division by zero. We actually determine limiting values of function at these points and exclude those values of y from the real set of y, which is determined assuming x belonging to R.

There are certain cases in which denominator of the rational function can not become zero. Consider rational functions :

f x = 2 x 2 − x + 1 x 2 + 1 f x = 2 x 2 − x + 1 x 2 + 1 g x = x + 1 2 x 2 − x + 1 g x = x + 1 2 x 2 − x + 1 h x = 2 x 2 − x + 1 | x | + 1 h x = 2 x 2 − x + 1 | x | + 1

The denominators of all these functions can not be zero. Under this condition, domain of the function is real number set R.

Example 3

Problem : Find the range of function :

f x = x 1 + x 2 f x = x 1 + x 2

Solution : The denominator of the given rational function can not be zero. Hence, domain of function is real number set R. There is no exclusion point. Rearranging to form a quadratic equation in x, we have :

⇒ y + y x 2 = x ⇒ y + y x 2 = x

⇒ y x 2 − x + y = 0 ⇒ y x 2 − x + y = 0

We should analyze for coefficient of “ x 2 x 2 ” in the quadratic equation. For quadratic equation, coefficient of “ x 2 x 2 ” can not be zero i.e. y ≠ 0. For real x, y ≠ 0 and D≥0 :

D = - 1 2 − 4 X y X y = 1 − 4 y 2 ≥ 0 D = - 1 2 − 4 X y X y = 1 − 4 y 2 ≥ 0 ⇒ y 2 ≥ 1 4 ⇒ y 2 ≥ 1 4 ⇒ y ∈ [ - 1 2 , 1 2 ] ⇒ y ∈ [ - 1 2 , 1 2 ]

What if y=0? Putting this value in the quadratic equation, we have :

⇒ 0 − x + 0 = 0 ⇒ 0 − x + 0 = 0

⇒ x = 0 ⇒ x = 0

This is included in the domain. Hence, y=0 is included in the range. The range of the rational function, therefore, remains unaffected :

⇒ y ∈ [ - 1 2 , 1 2 ] ⇒ y ∈ [ - 1 2 , 1 2 ]

Example 4

Problem : Find the range of the function :

y = f x = x 2 − 5 x + 4 x 2 − 3 x + 2 y = f x = x 2 − 5 x + 4 x 2 − 3 x + 2

Solution : We see that discrimanants of numerator and denominator polynomials are positive. On factorizing,

⇒ y = x 2 − 5 x + 4 x 2 − 3 x + 2 = x − 1 x − 4 x − 1 x − 2 ⇒ y = x 2 − 5 x + 4 x 2 − 3 x + 2 = x − 1 x − 4 x − 1 x − 2

Clearly, rational function is not defined for x=1 and x=2. Domain of the function is R- {1,2). For the sake of determining range, the limiting values of function for these values of x are obtained by canceling (x-1) from numerator and denominator :

⇒ y = x − 4 x − 2 ⇒ y = x − 4 x − 2

For x=1, y = 3. For x=2, however, the function value is indeterminate. In totality, we need to exclude y=3 from the interval of real values of y. Now, in order to determine real values of y, we rearrange the given function to form a quadratic equation in x :

⇒ y x 2 − 3 y x + 2 y = x 2 − 5 x + 4 ⇒ y x 2 − 3 y x + 2 y = x 2 − 5 x + 4 ⇒ y − 1 x 2 + 5 − 3 y x + 2 y − 4 = 0 ⇒ y − 1 x 2 + 5 − 3 y x + 2 y − 4 = 0

We should analyze for coefficient of “ x 2 x 2 ” in the quadratic equation. For quadratic equation, coefficient of “ x 2 x 2 ” can not be zero i.e. y-1 ≠ 0. For real x, y-1 ≠ 0 and D≥0.

For y-1 = 0, y = 1. Putting this value in the quadratic equation,

⇒ 0 + 5 − 3 x + 2 − 4 = 0 ⇒ 0 + 5 − 3 x + 2 − 4 = 0 ⇒ x = 1 ⇒ x = 1

We see that x=1 is not part of domain. This is actually the value which reduces denominator to zero. Hence, we should exclude y = 1 from the real values of y. Now for D≥0,

D = 5 − 3 y 2 − 4 y − 1 2 y − 4 ≥ 0 D = 5 − 3 y 2 − 4 y − 1 2 y − 4 ≥ 0 ⇒ 25 + 9 y 2 − 30 y − 4 { 2 y 2 − 6 y + 4 } ≥ 0 ⇒ 25 + 9 y 2 − 30 y − 4 { 2 y 2 − 6 y + 4 } ≥ 0 ⇒ 25 + 9 y 2 − 30 y − 4 { 2 y 2 − 6 y + 4 } ≥ 0 ⇒ 25 + 9 y 2 − 30 y − 4 { 2 y 2 − 6 y + 4 } ≥ 0

The coefficient of y 2 y 2 is positive. The discriminant is 0. Clearly, following sign rule, f(x) ≥0 for all real values of y. Hence, real values of y are real number set R. However, we need to exclude y = {1,3) as discussed above. Therefore, range of given function is R-{1,3}.

Alternative

Once, exception points are noted, we can evaluate “y” from the reduced form :

⇒ y = x − 4 x − 2 ⇒ y = x − 4 x − 2

Solving,

⇒ x = 2 y - 4 y − 1 ⇒ x = 2 y - 4 y − 1

Clearly, y#1. But we have seen that y#3 as well. Hence, range of rational function is R-{1,3}.

Graph of rational function

We know that rational function is a composition of two functions in the following form,

f x = p x q x f x = p x q x

where q(x) ≠ 0. If q(x) = 0, then the ratio has the form “ x / 0 x / 0 ”, which is not defined.

For plotting, let us consider a simple rational function given by, f x = 1 / x f x = 1 / x . This function is known as reciprocal function. It is not defined for x = 0. In order to plot the function, we calculate few initial values as :

F o r x = - 1, y = - 1 F o r x = - 1, y = - 1

F o r x = - 2, y = - 0.5 F o r x = - 2, y = - 0.5

F o r x = - 3, y = - 0.33 F o r x = - 3, y = - 0.33

F o r x = 0, y is not defined F o r x = 0, y is not defined

F o r x = 1, y = 1 F o r x = 1, y = 1

F o r x = 2, y = 0.5 F o r x = 2, y = 0.5

F o r x = 3, y = 0.33 F o r x = 3, y = 0.33

The graph of the function is shown here :

Figure 1: This plot is not defined at x = 0.
Rational function
Rational function (rf4.gif)

This plot is not defined at x = 0. The domain of the given function, therefore, is real numbers, “R” except zero. Also,

⇒ x = 1 y ⇒ x = 1 y

This means that function value can not be zero. Hence, range of the function is also real numbers, “R” except zero.

D o m a i n = R − { 0 } D o m a i n = R − { 0 }

R a n g e = R − { 0 } R a n g e = R − { 0 }

Example 5

Problem : Draw the graph of rational function given by :

f x = x 2 − 1 x − 1 f x = x 2 − 1 x − 1

Discuss the nature of graph and also determine domain and range of the given function.

Solution : The form of the given function is that of rational function. We observe that the function is not defined for "x = 1" as function has the form " x / 0 x / 0 ", which is undefined. The domain of the given function, therefore, is “R” except “1”. It should be noted that while interpreting domain or range we should not cancel out common terms in the numerator and denominator.

For other values of “x”, the value of the function is given by the reduced expression :

f x = x 2 − 1 x − 1 = x + 1 f x = x 2 − 1 x − 1 = x + 1

Clearly, if the given function were valid for x =1, then y = x+1 = 1 + 1 = 2. Thus, function f(x) can take any real value except “2”. Hence, range of the function is "R" except "2". The domain and range of the given function are :

D o m a i n = R − { 1 } D o m a i n = R − { 1 }

R a n g e = R − { 2 } R a n g e = R − { 2 }

In order to plot the function, we calculate few initial values as :

F o r x = - 3, y = − 2 F o r x = - 3, y = − 2

F o r x = - 2, y = − 1 F o r x = - 2, y = − 1

F o r x = - 1, y = 0 F o r x = - 1, y = 0

F o r x = 0, y = 1 F o r x = 0, y = 1

F o r x = 1, y is not defined F o r x = 1, y is not defined

F o r x = 2, y = 3 F o r x = 2, y = 3

F o r x = 3, y = 4 F o r x = 3, y = 4

The graph of the function is shown here :

Figure 2: The plot is not defined at x = 1.
Rational function
Rational function (rf5.gif)

The plot is not defined at x = 1. There is a break at x = 1.

Nature of graph

Here, we consider graphs of rational functions of type :

y = 1 x , 1 x 3 , 1 x 5 ,............. y = 1 x , 1 x 3 , 1 x 5 ,.............

The nature of graph of these rational function of type y = 1 x n y = 1 x n , where n is an odd integer such that n≥ 1, is similar to graph of y=1/x as shown in the figure. The graph is that of rectangular hyperbola.

Figure 3: Graph of rational function.
Rational function
Rational function (r5.gif)

We need to emphasize that the graph generalizes the nature and is helpful to estimate domain and range of functions. We need to graph individual function if required. The nature of graph of function type y = 1 x n y = 1 x n , where n is an even integer such that n≥2 is shown in the figure below :

y = 1 x 2 , 1 x 4 , 1 x 4 ,............. y = 1 x 2 , 1 x 4 , 1 x 4 ,.............

Figure 4: Graph of rational function.
Rational function
Rational function (r6.gif)

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